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Determining when a function is an element of L^2 (0,1)

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine for what k f(x)=xk is an element of L2 (0,1) vector space
    k ∈ ℝ

    2. Relevant equations


    3. The attempt at a solution
    [tex] \int_{0}^{1} x^{2k} dx = \frac{1-0^{2k+1}}{1+2k} = \sum_{n=0}^{\infty}{(-2k)^{n}} [/tex] (for k > -½)

    This sum should converge for [tex]
    \lim_{n \to +\infty}
    {\frac{|(-2k)^{n+1}|}{|(-2k)^{n}|}} < 1
    =
    |-2k| < 1
    [/tex]

    Which gives me a radius of convergence for
    [tex] - \frac{1}{2} < k < \frac{1}{2}
    [/tex]

    But just by examining it, the integral should exist for any k greater than negative one-half, what is wrong with my ratio test?
     
    Last edited: Oct 17, 2014
  2. jcsd
  3. Oct 17, 2014 #2

    LCKurtz

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    The infinite series doesn't represent the fraction for all values of k. The series may not converge for some k but that doesn't mean the fraction's value doesn't exist.
     
  4. Oct 17, 2014 #3
    Ok thanks, obvious mistake
     
  5. Oct 18, 2014 #4

    HallsofIvy

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    I don't see your logic. The integral of [itex]x^{2k}[/itex] is the number 1/(2k+ 1) as long as that number exists. What does that have to do with whether or not there is a geometric sequence converging to it? If k= 1, which is greater than 1/2, f(x)= x which is certainly twice integrable between 0 and 1.
     
  6. Oct 18, 2014 #5
    someone already responded pointing out it was wrong in an actually helpful way, and I already responded back, so move along now
     
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